1 Answer
1

Let $g$ be the identification
\begin{align}
g(0) = (1,0,0) &\quad g(5) = (-1,0,0) \\
g(1) = (0,1,0) &\quad g(4) = (0,-1,0) \\
g(2) = (0,0,1) &\quad g(3) = (0,0,-1).
\end{align}
Then $f$ is given by
$$ f(m,n) = g^{-1}( g(n)\times g(m) ) ,$$
so under the identifications given by $g$ your map is just the vector product (in reverse order because of an unfortunate difference in convention between the vector product and what you want your map to be).
Is this elegant enough?